Viewing 14 posts - 1 through 14 (of 14 total)
  • One for the wheelbuilders…
  • cynic-al
    Free Member

    …or members of the Institute of Physics 😛

    Of course there are many factors, but will a heavily dished wheel be 1. stronger 2. stiffer with 12 spokes a side or 16 DS and 8 NDS? This on a stans rim – requiring low tension.

    I suspect the answer is 1. no 2. yes

    coffeeking
    Free Member

    Interesting question. Not sure. Won’t be stiffer, I’m 99% certain (see earlier thread on wheel stiffness).The question is how does the wheel fail. Generally I think they fail due to passing the “0” angle on the DS spokes and that allowing the collapse of the tension in both sides but I’ve not really thought that through so I can’t be sure. I think you’d be better off having the wheel less dished in the frame, tyre space allowing.

    cynic-al
    Free Member

    Are you 1. a wheelbuilder? 2. a member of IOP?

    stuey
    Free Member

    C-al is it heavily dished on both sides – or offset like a pugsley?

    Elfinsafety
    Free Member

    Ooh! I’m a member of the Ponds Institute, Al! 😀

    cynic-al
    Free Member

    I know you are Pond life Fred 😛

    stuey – dish is dish, but it’s not offset.

    Elfinsafety
    Free Member

    😆

    (Cackles with glee at Al’s rudeness)

    TheBrick
    Free Member

    Ok simple model here but here goes.

    Take the angle between the NDS spoke and the vertical to be theta _1.
    Take the angle between the NS spoke and the vertical to be theta _2.
    The tension in the NDS spoke to be T_1.
    The tension in the DS spoke to be T_2.
    (Draw it)

    Considering lateral components on a wheel with equal number of spokes.

    T_1 sin theta_1 = T_2 sin theta_2,

    sin theta_1/sin theta_2 = T_2/T_1
    =R_1,

    the ratio between the tensions.
    On a dished wheel theta_2 < theta_1 => sin theta_1 > sin theta_2 => r_1 > 1. Which we already know.

    Now consider the case where there are a unequal number of spokes, we shall donate this by a ratio s_r where
    s_r = n_1/n_2,
    where n_1 = number of spokes on NDS and n_2 = number of spoke on drive side. Again considering the lateral components

    s_r T’_1 sin theta_1 = T’_2 sin theta_2

    s_r sin theta_1/sin theta_2 = T’_2/ T’_1
    =R_2
    = s_r R_1.

    For the wheel with an unequal number of spokes on either side to be stronger.

    |1-s_r R_1| < |1-R_1|

    although we know R_1 > 1 we don’t know if s_r R_1 > 1 hence the absolute value is needed.

    Squaring

    1 – 2s_r R_1 +s_2^2 R_1^2 < 1- 2R_1 + R_1^2

    (1-s_r^2)R_1<2(1-s_r)

    R_1 < 2(1-s_r)/(1-s_r^2).

    In your case

    s_r = 8 / 16 = 1 / 2

    so for this to be stronger than an equal number of spokes

    R_1 < 2(1/2)/(3/4)=4/3

    =>sin theta_1 < 4/3 sin theta_2.

    So if the depending on the amount of dish, maybe, laterally, but radially might be a different case.

    Elfinsafety
    Free Member

    stuey
    Free Member

    It’s hard not to agree with TheBrick. 😉

    GavinB
    Full Member

    Yup, glad you gave us the ‘simple model’ there TheBrick… 😯

    cynic-al
    Free Member

    I think we need aracer and Mark’s input on thebrick’s theories.

    I mean…formulas and stuff….WTF!!!

    aracer
    Free Member

    I’m not a member of the IoP, and only build wheels for my own amusement, so couldn’t possibly comment.

    takisawa2
    Full Member

    Be interesting to model this up & see what Ansys makes of it. Would take some serious setting up though. And would probably induce sleep. Quickly.

Viewing 14 posts - 1 through 14 (of 14 total)

The topic ‘One for the wheelbuilders…’ is closed to new replies.